Database Design and

The Entity-Relationship ModelLecture 21

R&G - Chapter 2

A relationship, I think, is like a shark, you know? It has to constantly move forward or it dies. And I think what we got on our hands is a dead shark.Woody Allen (from Annie Hall, 1979)

Administriva

• Homework 4 Due One Week From Today

• Midterm is graded

Exam Summary

• Average: 74.5

• Max: 93.5• Min: 36.5• Std Dev: 11.6

-20

-15

-10

-5

0

5

10

15

20

-40 -30 -20 -10 0 10 20 30

Diff from Mean Exam 2

Diff

from

Mea

n Ex

am 1

Good on Exam 1Worse on Exam 2

Bad on Exam 1Good on Exam 2

Exam Details

Section: Low – High / Total (Average)• 1: External Sorting 10 – 24 / 24 (18)• 2: Query Execution 17 – 43 / 44 (36)• 3: Query Optim. 6.5 – 29 / 32 (20)

Section 1: External Sorting

• For all questions in this section, assume that you have 96 pages in the buffer pool, and you wish to sort EMP by Ename.

• Using the general external sort, how many sorting passes are required? (6pts)– #passes = 1 + Ceil(log(B-1)(N/B))– = 1 + Ceil(log(95)(10,000/96))– = 1 + 2 – = 3

• What is the I/O cost of doing this sort? (6pts)– I/O Cost = 2N * (# passes)– = 20,000 * 3

Section 1 (cont)• Database systems sometimes use “blocked I/O” because reading a

block of continuous pages is more efficient that doing a separate I/O for each page. Assume that in our computer system, it is much more efficient to read and write blocks of 32 pages at a time, so all reads and writes from files must be in blocks of 32 pages (and if a file has less than 32 pages, it is padded with blank pages). Consider doing external sort with blocked I/O, which gets faster I/O at the expense of more sorting passes. If the database must read and write 32 pages at a time, how many sorting passes are required? Show either the formula you use, or the temp file sizes after each pass. (4pts)

– Pass 0: 105 runs of 96 pages– Pass 1: 2-way merge 96 + 96 -> result is 53 runs of 192 pages– Pass 2: 2-way merge of 192 + 192 -> result is 27 runs of 384 pages– Pass 3: -> result is 14 runs of 768 pages, – Pass 4: -> result is 7 runs 1536 pages, – Pass 5: -> result is 4 runs 3072 pages, – Pass 6: -> result is 2 runs of 6144, – Pass 7: -> result is 10,000– 8 passes total

Section 1 (cont)• Blocked I/O is used because reading a 32-page block

is faster than 32 separate 1-page I/Os. For each sorting pass, you still must read and write every page in the file, but instead of doing 10,000 1-page I/Os, instead you do (10,000/32) block I/Os. Assume that in our system, we can read a 32-page block in the time it would normally take to do 8 single-page I/Os. Is the blocked I/O sort faster or slower than a regular sort from question 2? By approximately what ratio? (4pts)

• Assume that instead of a heap file, the records from EMP are stored a clustered B-Tree index, whose key is Ename, using alternative 1 (i.e., the full data records are stored in the leaves of the tree). The B-tree has depth 3. Assuming the B-Tree already exists, what is the approximate I/O cost to use the B-tree to get the records in sorted order? (4pts)

Section 2: Relational Operators• Consider the operation: (EID < 5000)EMP

• What is the I/O cost of this operation if there is no index? (4pts)– Lecture 15 slide 9: no index -> sequential scan, N I/Os, meaning

10,000 I/Os

• What is the reduction factor? (4pts)– 5000/100,000 = 0.05

• Assume that instead of a heap file, the records from EMP are stored in a clustered B-Tree index, whose key is Ename, using alternative 1 (i.e., the full data records are stored in the leaves of the tree). The B-tree has depth 3. Assuming the B-Tree already exists, what is the I/O cost of this selection operation? (4pts)– Due to a copy paste error on my part, the B-Tree Index is *not*

useful for the query. I gave credit either way, whether people tried to use the index as an index, or for sequential scan.

– In considering the cost of using the index, either as an index or for a scan, full credit required knowing that B-Trees have an overhead of approximately 50%, i.e., there are 50% more leaves than the number of pages in a heap file.

Relational Operators (cont)

• Consider the query: select distinct Ename from EMP order by Ename– If you had 96 pages of memory, what algorithm

would you use to execute this query and why? (4pts)

– Order by suggests sort-merge to remove dups.

• If, instead, the query were: select distinct Ename from EMP and you had 101 pages of memory, what algorithm would you use and why? (4pts)– Could use hashing to remove dups, which would

have a similar I/O cost but be more CPU efficient.

Relational Operators (cont)• Consider the join: In_Dept Dept

– Assuming 100 pages of memory, what is the I/O cost of this join using Blocked Nested Loops? (4pts)

– Lec 15, slide 27: Blocked nested loops: M + Ceil(M/(B-2)) * N– 550 + Ceil(550/100)) * 20 = 550 + 6 * 20 = 670

• What is the I/O cost of this using Index Nested Loops, with an unclustered Hash index on Dept.DID? Assume an average cost of 1.2 I/Os to get to the right hash bucket. (4pts)

– Lec 15, slide 27: index nested loops: M + #tuples in M * index cost– 550 + 110,000 * (1.2 + 1) = 242,000

• Consider the join: Dept In_Dept. Assuming 100 pages of memory, what is the I/O cost of this using Sort-Merge-Join, optimized so that the last merge pass is combined with the join if possible? (4pts)

– Best: 1 pass over In_Dept, to make 6 runs, sort Dept in memory, merge. Cost 550*3 + 20 = 1670

– By the book: 3(M + N) = 3*570 = 1710

• Assuming 100 pages of memory, what is the I/O cost of this using regular (not hybrid) hash join? (4pts)

– Best: partition In_Dept (2 * 550), hash Dept in memory, match. Cost 550 * 3 + 20 = 1670

– By the book: 3(M + N) = 1710

Relational Operators (cont)• Consider the join: ( (EID <= 100)EMP) In_Building Assume

there is no index on EMP, and an unclustered hash index on In_Building.EID, and plan operators are pipelined where possible. You have 100 pages of memory. Assume that it takes on average 1.2 I/Os get reach the appropriate hash bucket.

• What is the I/O cost of this using Blocked Nested Loops? (4pts)– M + Ceil(M/(B-2)) * N – If you do not push select down:10,000 + Ceil(10,000/(100-2)) * 550

= 10,000 + 103*550 = 66,650– If you do push select, pipeline result: Size of selection in pages is S

= (I/100,000)*10,000 = I/10 = 10– Join cost: 10,000 + Ceil(10/(100-2)) * (550) = 10,550

• What is the I/O cost of this using Index Nested Loops? (4pts)– If you do push select, pipeline result: Size of selection in pages is:– S = (I/100,000)*10,000 = I/10 = 10– each EID will have 1.1 matching tuples. – Join cost: M + (# tuples * index cost)– 10,000 + 100 * (1.2 + 1.1) = 10,230

Section 3: Query Optimization• There are several parts of an SQL query that can cause a column (or

columns) to be considered an "interesting order", meaning that an ordered input can lower the cost of evaluating that part of the query. Briefly describe three of these, and for each one briefly explain why. (6 pts)– Order By, Group By, Joins, Distinct, some aggregations (e.g. Max, Min)

• Write the reduction factor that would be used by a System R style query optimizer for each of the following predicates (6pts)– Building.BID > 150

• BID ranges from 1 to 200, so this would be ¼.– EMP.Ename = “Joe”

• We don’t know about distribution of names, so 1/10.– IN_DEPT.EID = 003

• Since there are 100,000 employees, this is 1/100,000

• Given the following query, where “X” is the join operator: π(EMP.Ename) (Dept.Budget > 500000) (Emp X In_Dept X Dept)

• Mark whether each of the following queries are equivalent (True/False). – __T___π(EMP.Ename) (Dept.Budget > 500000) (Emp X Dept X In_Dept)– __T___π(EMP.Ename) (Dept.Budget > 500000) (Dept X In_Dept X Emp)– __T___π(EMP.Ename) (Emp X In_Dept X (Dept.Budget > 500000)(Dept))– __F___(Dept.Budget > 500000) (π(EMP.Ename) (Emp) X In_Dept X Dept)

Query Optimization (cont)• Given the schema on the first page of the exam, assume the following:

– There are unclustered hash indexes on both Emp.EID and Dept.DID.– There are clustered B-Tree Indexes (alt 1 – data records store in the B-Tree

leaves) on both Emp.EID and Dept.DID.– There is an unclustered Btree index on (Emp.Salary, Emp.EID)– You can assume that Btree-Indexes have heights of 3. And the cost for

getting to the data record using a hash index is 2.

• What is the best access plan for the following query (4pts).SELECT E.eid, E.SalaryFROM Emp EWHERE E.Salary > 100,000– Very best is unclustered B-tree, index only plan

• Draw all possible join orders considered by a System-R style query

optimizer for the following query. (4pts)SELECT E.eid, D.dnameFROM Emp E, In_Dept I, Dept DWHERE E.sal = 64,000 AND D.budget > 500,000 AND E.eid = I.eid AND I.did = D.did– (E X I) X D)– (I X E) X D)– (D X I) X E)– (I X D) X E)

Where are we?

• This week: Database Design (me)– The ER Model as a design tool

• Next Week: Concurrency Control & Recovery (Prof. Roth)

• Week after: More Database Design (me)

Today and Thursday: The ER Model

• A different data model from Relational

• Most commonly used for database design

• Today: Details of the ER Model

• Thursday: Translating ER Schemas to Relational

Databases Model the Real World• “Data Model” translates real world things

into structures computers can store• Many models:

– Relational, E-R, O-O, Network, Hierarchical, etc.

• Relational– Rows & Columns– Keys & Foreign Keys to link Relations

sid name login age gpa53666 Jones jones@cs 18 3.453688 Smith smith@eecs 18 3.253650 Smith smith@math 19 3.8

sid cid grade53666 Carnatic101 C53666 Reggae203 B53650 Topology112 A53666 History105 B

Enrolled Students

Database Design

• The process of modelling things in the real world into elements of a data model.

• I.E., describing things in the real world using a data model.

• E.G., describing students and enrollments using various tables with key/foreign key relationships

• The Relational model is not the only model in use…

A Problem with the Relational Model

With complicated schemas, it may be hard for a person to understand the structure from the data definition.

CREATE TABLE Students(sid CHAR(20), name CHAR(20), login CHAR(10), age INTEGER, gpa FLOAT)

CREATE TABLE Enrolled(sid CHAR(20), cid CHAR(20), grade CHAR(2))

sid name login age gpa53666 Jones jones@cs 18 3.453688 Smith smith@eecs 18 3.253650 Smith smith@math 19 3.8

Studentscid grade sid Carnatic101 C 53666 Reggae203 B 53666 Topology112 A 53650 History105 B 53666

Enrolled

One Solution: The E-R Model• Instead of relations, it has:

Entities and Relationships

• These are described with diagrams, both structure, notation more obvious to

humans

• A visual language for describing schemas

lotname

Students

ssn

Enrolled_in

sincedname

budgetdid

Courses

Steps in Database Design• Requirements Analysis

– user needs; what must database do?• Conceptual Design

– high level descr (often done w/ER model)• Logical Design

– translate ER into DBMS data model• Schema Refinement (in 2 weeks)

– consistency, normalization• Physical Design (discussed already)

– indexes, disk layout• Security Design

– who accesses what, and how

Conceptual Design• Define enterprise entities and relationships

• What information about entities and relationships should be in database?

• What are the integrity constraints or business rules that hold?

• A database `schema’ in the ER Model is represented pictorially (ER diagrams).

• Can map an ER diagram into a relational schema.

ER Model Basics

• Entity: Real-world thing, distinguishable from other objects. Entity described by set of attributes.

• Entity Set: A collection of similar entities. E.g., all employees. – All entities in an entity set have the same set of

attributes. (Until we consider hierarchies, anyway!)– Each entity set has a key (underlined).– Each attribute has a domain.

Employees

ssnname

lot

ER Model Basics (Contd.)

• Relationship: Association among two or more entities. E.g., Attishoo works in Pharmacy department.– relationships can have their own attributes.

• Relationship Set: Collection of similar relationships.– An n-ary relationship set R relates n entity sets E1 ... En ;

each relationship in R involves entities e1 E1, ..., en En

lotname

Employees

ssn

Works_In

sincedname

budgetdid

Departments

ER Model Basics (Cont.)

• Same entity set can participate in different relationship sets, or in different “roles” in the same set.

subor-dinate

super-visor

Reports_To

since

Works_In

dnamebudgetdid

Departments

lot

name

Employees

ssn

Key Constraints An employee

can work in many departments; a dept can have many employees.

1-to-11-to ManyMany-to-Many

since

Manages

dname

budgetdid

Departments

since

Works_In

lot

name

ssn

Employees

In contrast, each dept has at most one manager, according to the key constraint on Manages.

Participation Constraints• Does every employee work in a department? • If so, this is a participation constraint

– the participation of Employees in Works_In is said to be total (vs. partial)– What if every department has an employee working in it?

• Basically means “at least one”

lotname dname

budgetdid

sincename dname

budgetdid

since

Manages

since

DepartmentsEmployees

ssn

Works_In

Means: “exactly one”

Weak EntitiesA weak entity can be identified uniquely only by

considering the primary key of another (owner) entity.– Owner entity set and weak entity set must participate in a

one-to-many relationship set (one owner, many weak entities).– Weak entity set must have total participation in this

identifying relationship set.

lot

name

agepname

DependentsEmployees

ssn

Policy

cost

Weak entities have only a “partial key” (dashed underline)

Binary vs. Ternary Relationships

If each policy is owned by just 1 employee: Bad design

Beneficiary

agepname

Dependents

policyid cost

Policies

Purchaser

name

Employees

ssn lot

Better design

• Think through all the constraints in the 2nd diagram!

Policies

policyid cost

agepname

DependentsCovers

name

Employees

ssn lot

Key constraint on Policies would mean policy can only cover 1 dependent!

Binary vs. Ternary Relationships (Contd.)

• Previous example illustrated case when two binary relationships were better than one ternary relationship.

• Opposite example: a ternary relation Contracts relates entity sets Parts, Departments and Suppliers, and has descriptive attribute qty. No combination of binary relationships is an adequate substitute.

Binary vs. Ternary Relationships (Contd.)

– S “can-supply” P, D “needs” P, and D “deals-with” S does not imply that D has agreed to buy P from S.

– How do we record qty?

Suppliers

qty

DepartmentsContractParts

Suppliers

Departments

deals-with

Parts

can-supply

VS.

needs

Summary so far

• Entities and Entity Set (boxes)• Relationships and Relationship sets (diamonds)

– binary– n-ary

• Key constraints (1-1,1-M, M-M, arrows on 1 side)

• Participation constraints (bold for Total)• Weak entities - require strong entity for key• Next, a couple more “advanced” concepts…